Here I have indicated all the pairs that are as yet unresolved with the "/" -- triples or quads are marked ***, except for a few (r1c3, r1c7, and r9c7) that will be important in the ensuing argument.

The next logical move works like this. A "5" can only occupy one of two possible cells in column 7. Similarly, there are only two possibilities for "5" in row 1, and two possibilities for "5" in row 8. Suppose that r9c7 = "5". Then we have the following chain of inference:

r9c7=5 ==> r1c3=5 ==> r8c3=8 ==> r8c9=5

But this can't be right, because it forces two "5"s into the same 3x3 box, in the lower right corner of the grid. So the "5" in column 7 must appear at r1c7.

The next move is also pretty tricky. Start with the {3, 8} pair at r6c3. Since "3" can only appear in two places in row 6, we have the following chain of inference (following across row 6, then down):

r6c3=8 ==> r6c9=3 ==> r8c9=5

But there's another chain to be traced out, because the value "8" can only appear in two places in column 3. Following this chain we see that

r6c3=8 ==> r8c3=5 ==> r8c9=3

But this is impossible -- we can't put two different values in the same cell! We conclude that r6c3 is not an "8", and therefore r6c3 = 3.

Hi David,
I like the way you think.
It is always easy to find the "complicate solution".
Now let us start with what is the XY-wing?
If we have:
XY XZ

YZ *

It can be easily seen that whichever value is in XY, the cell marked with the asterisk cannot be Z.
if XY = X, then XZ = Z, so * cannot be Z
if XY = Y, then YZ = Z, so * cannot be Z
This allows Z to be eliminated from the candidates for the marked cell.

Doesn't it look simple and easy to detect ?

Now lets take a look at the position you have got.
It is the G-point!

3 can't be in r6c9, because we have XY-Wing numbers
X=5 Y=8 in r8c3
X=5 Z=3 in r8c9
Y=8 Z=3 in r6c3

And this is enough. The rest of the numbers are comming by themself before the whisky gets warm and the coffee cold.

It was a pleasure to meet u. This one you killed before Katie.
Would you like an other XY-wing? Of course with 17 cells.

The algebraic notation used in explanation can be off-putting and so I
have attempted to distil a simple description of when XY-Wing applies.

Consider a rectangle of four cells (however small or large) using the
four corner cells. If three of the corners from a triplet of values with
two occurrences of each value, the fourth corner CANNOT contain the
digit value that does not occur in its opposite corner.

Example:

18; 16
68; 68
There is a triplet {18,16,68} in three corners (left and top) so that the
bottom right cannot contain 6 (the value not in 18 opposite).
ALSO
There is a triplet {18,16,68} in three corners (RIGHT and top) so that the
bottom LEFT cannot contain 8 (the value not in 16 opposite).

This resolves to

18; 16
6,8
and further progress will depend on non-quoted cells but should be
very much easier.

+++

This rule above applies only when three of the four corners have been
reduced to pairs. Does the rule have wider scope and apply when more
than two digits are in any of the three corners? It would seem that the
number of digits in the fourth (target) corner is not material.

NB: Other readers of this forum should check for themselves that the
suggested rule works before including it in their repertoire! You have been
warned, and only practice will confirm usefulness.